91Ó°ÊÓ

Consider a long, solid cylinder that has a circular cross-section of radius a and is oriented with its axis in the z-direction, the current density is not constant across the cross-section of the wire and it is given by

$\stackrel{→}{J}=\left(\frac{b}{r}\right)e^{(r−a)/\mathrm{δ}\widehat{k}}$ …â¶Ä¦(1)

where r is the distance from the center. First, we need to find an expression for the total current passing through the entire cross-section of the wire $I_o$in terms of b, $\mathrm{δ}$, and a. Consider a small segment, a ring with a width of dr and at distance r from the center of the circle, the current in this element equals the area of the element multiplied by the current density J, that is,

dI = JdA

the area of the element equals the area of a rectangle with a width of dr and length of $2\mathrm{Ï€}r$(the circumference of a circle with a radius of r), that is,

localid="1668267631483" $dA=2\mathrm{Ï€°ù»å°ù»å}$ (2)

Thus,

localid="1668267634906" $dI=\left(\frac{b}{r}\right)e^{(r−a)/\mathrm{δ}}2\mathrm{π}rdr=2\mathrm{π}b{e}^{(r−a)/\mathrm{δ}}dr$

the total current passing through the wire equals the integration over the whole wire, from r = 0 to r = a, that is,

localid="1668267641935" $\begin{array}{c}{I}_0={∫}_0^a dI=2\mathrm{π}b{∫}_0^a e^{\frac{r−a}{\mathrm{δ}}}dr\\ ={\left.2\mathrm{π}b\mathrm{δ}{e}^{\frac{r−a}{\mathrm{δ}}}\right|}_0^a=2\mathrm{π}b\mathrm{δ}\left(1−e^{\left.−\frac{a}{\mathrm{δ}}\right)}\right.\\ I_0=2\mathrm{π}b\mathrm{δ}\left(1−e^{−\frac{a}{\mathrm{δ}}}\right)\end{array}$

the radius of the cylinder is a = 5.00 cm, b is a constant equal to 600 A/m, and d is a constant equal to 2.50 cm, substitute with the givens to get,

localid="1668267664757" $\begin{array}{c}{I}_0=2\mathrm{π}(600\mathrm{A}/\mathrm{m})\left(0.025\mathrm{m}\right)\left(1−e^{(0.050/0.025)}\right)=81.5\mathrm{A}\\ I_0=81.5\mathrm{A}\end{array}$

Now for $râ‰\yen$a, the enclosed current is Io, therefore Ampere's law ($\text{∮}\stackrel{→}{B}â‹\dots \stackrel{→}{d}l={\mathrm{μ}}_o{I}_{\text{encl}}$) gives, $\text{∮}\stackrel{→}{B}â‹\dots \stackrel{→}{dl}=B2\mathrm{Ï€}r={\mathrm{μ}}_o{I}_{\text{encl}}={\mathrm{μ}}_o{I}_o$

Or,

$B=\frac{{\mathrm{μ}}_{\mathrm{ο}}{I}_{\mathrm{ο}}}{2\mathrm{π}r}$

Now we need to find the enclosed current for $r≤a$. Consider a ring with a width of dr and area of $dA=2\mathrm{Ï€°ù»å°ù»å}$, the enclosed current in this area is,

dI = JdA

Or,

$dI=2\mathrm{π}b{e}^{(r−a)/\mathrm{δ}}d{r}^{\mathrm{′}}$

integrate from r'=0 to r'= r to get,

$\begin{array}{c}dI=2\mathrm{π}b{∫}_0^r e^{(r−a)/\mathrm{δ}}dr={\left.2\mathrm{π}b\mathrm{δ}{e}^{\left(r^{\mathrm{′}}−a\right)/\mathrm{δ}}\right|}_0^r\\ =2\mathrm{π}b\mathrm{δ}\left(e^{(r−a)/\mathrm{δ}}−e^{−a/\mathrm{δ}}\right)=2\mathrm{π}b\mathrm{δ}{e}^{−a/\mathrm{δ}}\left(e^{r/\mathrm{δ}}−1\right)\end{array}$

But $I_0=2\mathrm{π}b\mathrm{δ}\left(1−e^{−a/\mathrm{δ}}\right)$, thus,

$I=I_0\frac{\left(e^{r/\mathrm{δ}}−1\right)}{\left(e^{a/\mathrm{δ}}−1\right)}$

Now we need to use the results of step 3 to find the magnetic field for $r≤a$. Ampere's law $\left(\text{∮}\stackrel{→}{B}â‹\dots \stackrel{→}{d}l={\mathrm{μ}}_o{I}_{encl}\right.$) gives

$\begin{array}{c}\text{∮}\stackrel{→}{B}â‹\dots d\stackrel{→}{l}=B\left(r\right)2\mathrm{Ï€}r={\mathrm{μ}}_0{I}_{\mathrm{encl}}={\mathrm{μ}}_0{I}_0\frac{\left(e^{r/}−1\right)}{\left(e^{a/\mathrm{δ}}−1\right)}\\ B=\frac{{\mathrm{μ}}_0{I}_0\left(e^{r/\mathrm{δ}}−1\right)}{2\mathrm{Ï€}r\left(e^{a/\mathrm{δ}}−1\right)}\end{array}$

Finally, we need to evaluate the magnitude of the magnetic field at $r=\mathrm{δ}$,r = a, and r=2a, as,

$\begin{array}{r}B(r=\mathrm{δ})=\frac{{\mathrm{μ}}_0{I}_0(e−1)}{2\mathrm{Ï€}\mathrm{δ}\left(e^{a/\mathrm{δ}}−1\right)}\\ =\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(81.5\mathrm{A}\right)}{2\mathrm{Ï€}\left(0.025\mathrm{m}\right)}\frac{(e−1)}{\left(e^{0.050/0.025}−1\right)}\\ =1.75×{10}^{−4}\mathrm{T}\\ B(r=\mathrm{δ})=1.75×{10}^{−4}\mathrm{T}\\ B(r=a)=\frac{{\mathrm{μ}}_0{I}_0}{2\mathrm{Ï€}a}\frac{\left(e^{a/\mathrm{δ}}−1\right)}{\left(e^{a/\mathrm{δ}}−1\right)}\\ =\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(81.5\mathrm{A}\right)}{2\mathrm{Ï€}\left(0.050\mathrm{m}\right)}\\ =3.26×{10}^{−4}\mathrm{T}\\ B(r=a)=3.26×{10}^{−4}\mathrm{T}\\ B(r=2a)=\frac{{\mathrm{μ}}_0{I}_0}{2\mathrm{Ï€}a}\\ =\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(81.5\mathrm{A}\right)}{2\mathrm{Ï€}\left(0.100\mathrm{m}\right)}\\ =1.63×{10}^{−4}\mathrm{T}\\ B(r=2a)=1.63×{10}^{−4}\mathrm{T}\end{array}$